Question

A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it?
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Answer 1

Explanation:

Radius of the hoop, r = 2 m

Mass of the hoop, m = 100 kg

Velocity of the hoop, v = 20 cm/s = 0.2 m/s

Total energy of the hoop = Translational K.E. + Rotational K.E.

ET = (1/2)mv2 + (1/2) I ω2

Moment of inertia of the hoop about its centre, I = mr2

ET = (1/2)mv2 + (1/2) (mr2)ω2

But we have the relation, v = rω

∴ ET = (1/2)mv2 + (1/2)mr2ω2

= (1/2)mv2 + (1/2)mv2 = mv2

The work required to be done for stopping the hoop is equal to the total energy of the hoop.

∴ Required work to be done, W = mv2 = 100 × (0.2)2 = 4 J.

Answer 2

Answer:

r=2m

m=100kg

speed of centre of speed=0.2m/s

W=F*S

ma*S

100*2*2

so work to be done=400J